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2 years ago

Which type of vehicle is used routinely at construction and mining sites?

Engineering

2 answers:

Semmy [17]2 years ago

6 0

Answer:

Hydraulics as the name implies is pertaining to water. Hydraulic vehicle is a vehicle used on the construction site. The vehicle makes use of liquids to function. The fluid powers the engine of the vehicle do as to generate power.

Types of fluids used in hydraulic vehicles includes: Water fluid, petroleum fluid, synthetic fluid.

Explanation:

it really depends on the job.

satela [25.4K]2 years ago

5 0

We are required to state the type of vehicle which is routinely used at the construction site.

The vehicle which is routinely used at the construction site is called hydraulic vehicle.

Hydraulics as the name implies is pertaining to water. Hydraulic vehicle is a vehicle used on the construction site. The vehicle makes use of liquids to function. The fluid powers the engine of the vehicle do as to generate power.

Types of fluids used in hydraulic vehicles includes: Water fluid, petroleum fluid, synthetic fluid.

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A 35-ft³ rigid tank has propane at 25 psia, 540 R and is connected by a valve to another tank of 20 ft³ with propane at 40 psia,

gulaghasi [49]

Answer:

final pressure = 200KPa or 29.138psia

Explanation:

The detailed step by step calculations with appropriate conversion factors applied are as shown in the attachment.

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3 years ago

Find the number of Btu conducted through a wall in 8 hours. The wall is 8 feet high by 24 feet long and has a total R-value of 1

dedylja [7]

Answer:

ΔQ = 4930.37 BTu

Explanation:

given data

height h = 8ft

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inside temperature t1 = 68°F

outside temperature t2 = 16°F

to find out

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solution

we get here number of Btu conducted by this expression that s

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put here value we get

$\frac{\Delta Q}{8} =\frac{-192}{16.2} (16-68)$

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ΔQ = 4930.37 BTu

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3 years ago

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Ksju [112]

Answer:

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2 years ago

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2 years ago

How to find the voltage(B Aab) in series parallel circuit?

Sindrei [870]

Answer:

Vab ≈ 3.426 V

Explanation:

First of all, it is convenient to find the equivalent parallel resistance of R5 and R6. That will be ...

R56 = (R5)(R6)/(R5 +R6) = (1000)(1500)/(1000 +1500) = 600

Then we can call V1 the voltage at the top of R2. The voltage at Va is a divider from V1:

Va = V1·(R4/(R3+R4)) = V1(560/1030) ≈ 0.543689V1

The voltage at Vb is also a divider from V1:

Vb = V1·(R7+R8)/(R2 +R56 +R7 +R8) = V1(780/1710) ≈ 0.456140V1

The parallel branches containing Va and Vb have an effective resistance of ...

(1030)(1710)/(1030+1710) = 642.81

That forms a divider with R1 to give V1:

V1 = (100 V)642.81/(1000 +642.81) ≈ 39.1287 V

The difference Va-Vb is ...

Vab = (39.1287 V)(0.543689 -0.456140) ≈ 3.426 V

_____

We have done this using parallel resistance and voltage divider calculations. You can also do it using node voltage equations. Using the same definition for V1 as above, we have ...

(Vs -V1)/R1 +(Vb -V1)/(R56+R2) +(Va-V1)/R3 = 0

(V1 -Vb)/(R56 +R2) -Vb/(R7+R8) = 0

(V1 -Va)/R3 -Va/R4 = 0

The solution of interest is the value of Vab, shown in the attachment. It computes as 154200/45013 V ≈ 3.42568 V.

4 0

3 years ago